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My pure amateur, possibly muddled understanding is that P=!NP is rather different from most open conjectures. A) It's extremely general or even "foundational"

My pure amateur, possibly muddled understanding is that P=!NP is rather different from most open conjectures. A) It's extremely general or even "foundational" in the sense that it's asking whether any algorithm at all exists to solve an extremely general sort of problem in polynomial time. B) All of the standard methods used to solve problems of this sort have at least been pronounced exhausted at this point. C) Very few theorems that inherently limit the speed of a class of problems actually have ever been proven. Very few methods for proving these constraints are known. D) A lot of famous theorems have yielded results through being embedded in a larger, different field where they are just one result of many proved with a new machinery (Fermet most prominently). But given P=!NP is so general it can't really embedded in a larger, tractible space - lots of things are equivalent to it but all these things are kind of the same.

someguy12342 7y ago
Great post. Here are a few things pointers if you want more information about your points: B) The situation is actually worse than exhausted. We know (have proven) that the regular methods simply won't work (check up "the reletaivzation barrier" and "natural proofs" and "algebrization") C) We actually can construct problems which take at least a certain amount of time to solve (though admittedly many of them are kind of unnatural). Search up "time hierarchy theorem". This is probably one of the theorems you mention. However, you're right, as far as natural problems go, it appears to be extreme difficult to prove most natural problems takes at least linear time (https://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np https://mathoverflow.net/questions/4953/super-linear-time-co...). Note the log*... factor grows extremely slowly. In fact, it is not known whether SAT requires more than linear time. Since SAT is NP-complete, we expect it to have no polynomial time algorithm... and yet here we are wondering whether it requires even more than linear time. If you are interested in this sort of thing, maybe look up fine-grained complexity. Also note that to solve P != NP, we will have to atleast shown P != PSPACE, since PSPACE contains NP. However, even this seemingly easier problem has had no progress and seems like it would require a huge breakthrough.
- joe_the_user 7y ago
  >>B) The situation is actually worse than exhausted. We know (have proven) that the regular methods simply won't work (check up "the reletaivzation barrier" and "natural proofs" and "algebrization") I've heard of those, even tried to read the proof. Still, my understanding is those claims are ultimately "informal proofs" despite having formal step so the situation isn't entirely closed-up but still very bad.
  - someguy12342 7y ago
    Maybe this will help with understand the revitalization barrier. Most proof of separations of complexity classes relativize, in that not only do they show A strict subset B they also show A^O strict subset B^O for ANY oracle O. Most of these proofs don't actually explicitly state this is true, but they are readily to be extended this way (e.g. the diagonlizing proofs of the time hierarchy theorem) However, we know (Baker, Gill, Solovay) that there exists and oracle O1 where P^O1 subset NP^O1 as well as an oracle O2 such that P^O2 not subset NP^O2 Therefore, we know we a proof of P != NP must not relativize or else that would contradict the aforementioned result.
    - pwdisswordfish2 7y ago
      P^O is always a subset of NP^O, whatever the O.
      - joe_the_user 7y ago
        I think the gp left out the strict before the later subset statements.
        someguy12342 7y ago
        That's right! I completely flubbed that part while doing some other edits! Thanks. What I wanted to say is there are O1 and O2 such that: P^O1 strict subset NP^O1 and P^O2 equals NP^O2
lacker 7y ago
It isn’t that different - most complexity theory conjectures are this way. Basically, there are all these classes of algorithm. P can be solved in polynomial time, NP can be solved in polynomial time by a computer that gets really lucky with rng, PSPACE can be solved with polynomial space, L can be solved in logarithmic space. For many of these pairs, it kinda feels like they are not the same, but we can’t prove it. So in particular we don’t know that P != PSPACE, which is a much weaker statement, and a very similar one, compared to P != NP. Similarly, we can’t prove there are cryptographically secure hash functions, and we can’t prove it’s hard to factor numbers. Basically, we just have very few mathematical tools for proving that complexity classes are different. It’s hard to prove that algorithmic problems are hard. We have diagonalization, which shows that the halting problem is impossible and that P != EXPTIME, and that’s about it. Anyway, I would conclude not that P != NP is a uniquely hard problem, as much as that complexity theory is very young compared to most fields of mathematics, and there are still many statements we can’t prove.
jeromebaek 7y ago
P vs NP is also distinguished in that many people think it has profound philosophical implications. Nick Land seems to think this (http://www.uf-blog.net/crypto-current-021/ http://www.uf-blog.net/crypto-current-021/), and so does Scott Aaronson (“if P=NP, the word would be a profoundly different place than we usually assume it to be ... there would be no value in “creative gaps””). Though I think Land’s idea is a lot more interesting — he says P vs NP is a paraphrase of the analytic vs synthetic distinction.

someguy12342 7y ago

Great post. Here are a few things pointers if you want more information about your points: B) The situation is actually worse than exhausted. We know (have proven) that the regular methods simply won't work (check up "the reletaivzation barrier" and "natural proofs" and "algebrization") C) We actually can construct problems which take at least a certain amount of time to solve (though admittedly many of them are kind of unnatural). Search up "time hierarchy theorem". This is probably one of the theorems you mention. However, you're right, as far as natural problems go, it appears to be extreme difficult to prove most natural problems takes at least linear time (https://mathoverflow.net/questions/4953/super-linear-time-complexity-lower-bounds-for-any-natural-problem-in-np https://mathoverflow.net/questions/4953/super-linear-time-co...). Note the log*... factor grows extremely slowly. In fact, it is not known whether SAT requires more than linear time. Since SAT is NP-complete, we expect it to have no polynomial time algorithm... and yet here we are wondering whether it requires even more than linear time. If you are interested in this sort of thing, maybe look up fine-grained complexity. Also note that to solve P != NP, we will have to atleast shown P != PSPACE, since PSPACE contains NP. However, even this seemingly easier problem has had no progress and seems like it would require a huge breakthrough.

joe_the_user 7y ago
>>B) The situation is actually worse than exhausted. We know (have proven) that the regular methods simply won't work (check up "the reletaivzation barrier" and "natural proofs" and "algebrization") I've heard of those, even tried to read the proof. Still, my understanding is those claims are ultimately "informal proofs" despite having formal step so the situation isn't entirely closed-up but still very bad.
- someguy12342 7y ago
  Maybe this will help with understand the revitalization barrier. Most proof of separations of complexity classes relativize, in that not only do they show A strict subset B they also show A^O strict subset B^O for ANY oracle O. Most of these proofs don't actually explicitly state this is true, but they are readily to be extended this way (e.g. the diagonlizing proofs of the time hierarchy theorem) However, we know (Baker, Gill, Solovay) that there exists and oracle O1 where P^O1 subset NP^O1 as well as an oracle O2 such that P^O2 not subset NP^O2 Therefore, we know we a proof of P != NP must not relativize or else that would contradict the aforementioned result.
  - pwdisswordfish2 7y ago
    P^O is always a subset of NP^O, whatever the O.
    - joe_the_user 7y ago
      I think the gp left out the strict before the later subset statements.
      - someguy12342 7y ago
        That's right! I completely flubbed that part while doing some other edits! Thanks. What I wanted to say is there are O1 and O2 such that: P^O1 strict subset NP^O1 and P^O2 equals NP^O2

lacker 7y ago

It isn’t that different - most complexity theory conjectures are this way. Basically, there are all these classes of algorithm. P can be solved in polynomial time, NP can be solved in polynomial time by a computer that gets really lucky with rng, PSPACE can be solved with polynomial space, L can be solved in logarithmic space. For many of these pairs, it kinda feels like they are not the same, but we can’t prove it. So in particular we don’t know that P != PSPACE, which is a much weaker statement, and a very similar one, compared to P != NP. Similarly, we can’t prove there are cryptographically secure hash functions, and we can’t prove it’s hard to factor numbers. Basically, we just have very few mathematical tools for proving that complexity classes are different. It’s hard to prove that algorithmic problems are hard. We have diagonalization, which shows that the halting problem is impossible and that P != EXPTIME, and that’s about it. Anyway, I would conclude not that P != NP is a uniquely hard problem, as much as that complexity theory is very young compared to most fields of mathematics, and there are still many statements we can’t prove.

jeromebaek 7y ago

P vs NP is also distinguished in that many people think it has profound philosophical implications. Nick Land seems to think this (http://www.uf-blog.net/crypto-current-021/ http://www.uf-blog.net/crypto-current-021/), and so does Scott Aaronson (“if P=NP, the word would be a profoundly different place than we usually assume it to be ... there would be no value in “creative gaps””). Though I think Land’s idea is a lot more interesting — he says P vs NP is a paraphrase of the analytic vs synthetic distinction.